Question

If A, B, C are the interior angles of a $\Delta$ABC, show that
$\cos \frac{B+C}{2}=\sin \frac{A}{2}$

Answer 1

Given, $\angle A,\angle B,and\angle C$ are interior angles of $△ABC$

$\angle A+\angle B+\angle C={180}^{\circ }$ -----------[i]

Consider,

$\cos \frac{(B+C)}{2}$

We know that,

$\cos \left(\theta \right)=\sin ({90}^{\circ }-\theta )$

$\therefore \cos \frac{(B+C)}{2}$

$=\sin ({90}^{\circ }-\frac{(B+C)}{2})$

$=\sin ({90}^{\circ }-\frac{({180}^{\circ }-A)}{2})$ ------------from[i]

$=\sin \left(\frac{{180}^{\circ }-({180}^{\circ }-A)}{2}\right)$

$=\sin \left(\frac{{180}^{\circ }-{180}^{\circ }+A}{2}\right)$

$=\sin \left(\frac{A}{2}\right)$

$=\sin \frac{A}{2}$

$\therefore \cos \frac{(B+C)}{2}=\sin \frac{A}{2}$